Question

Will the following molecules be optically active? (i) CH3CHOHCOOH (ii) CH2ClF Justify your answer on the basis of symmetry considerations.

Answer 1

Answer on Question #85387 – Chemistry – Other

Task:

Will the following molecules be optically active? (i) $\mathrm{CH}_3\mathrm{CH(OH)COOH}$ (ii) $\mathrm{CH}_2\mathrm{ClF}$. Justify your answer on the basis of symmetry considerations.

Solution:

(i):

optically active compound

2-hydroxypropanoic acid

$(S)$-2-hydroxypropanoic acid

$(R)$-2-hydroxypropanoic acid

The carbon atom (*) in 2-hydroxypropanoic acid carries four different substituents: H, OH, CH₃, and COOH. As a result, this molecule is chiral and it forms enantiomers.

(ii):

holorofluoromethane

not optically active compound

The carbon atom in chlorofluoromethane contains two identical H substituents. As a result, this compound is achiral and does not form enantiomers.

Answer: (i) will be optically active; (ii) will not optically active.

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